= Solution
For the prograde captured orbit, the departing half has instantaneous <velocity> tangent to the original orbit, in the original direction, with speed
$$
\boxed{u_{\rm depart}=2\sqrt{v^2+\frac{2GM}{r_0}}-\sqrt{\frac{GM}{r_0}}.}
$$
It exceeds the local <escape speed>: since $V\geq\sqrt2u_c$, $u_{\rm depart}\geq(2\sqrt2-1)u_c>\sqrt2u_c$. This is its speed immediately after the explosion, not its later speed at infinity.
If a counterrotating circular captured orbit is allowed, its signed velocity is $-u_c$, the outgoing speed is instead $2V+u_c$, and the required generated energy is $m(V+u_c)^2/2$. The usual answer takes the prograde orbit; specifying that choice removes the directional ambiguity.
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